3 free questions from this chapter (the full set has 30). Marking: +4 for correct · −1 for incorrect · 0 for unattempted.
Question 1Level 3 · AdvancedSingle correct
A block rests on the rough inclined face of a wedge of angle 37∘; the coefficient of static friction between them is 0.5. The wedge is given a horizontal acceleration a in the direction that presses the block against the inclined face. What is the largest value of a for which the block stays at rest relative to the wedge? Take g=10m/s2, sin37∘=0.6.
A1120m/s2
B7.5m/s2
C10m/s2
D20m/s2
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Answer: D
In the wedge frame a pseudo force ma acts horizontally, pushing the block into the face. At the upper limit the block tends to slide up, so friction acts down the incline.
Perpendicular to face: N=mgcosθ+masinθ. Along face: macosθ=mgsinθ+μN.
a=gcosθ−μsinθsinθ+μcosθ=10⋅0.8−0.30.6+0.4=20m/s2.
Traps: 20/11 is the lower limit (block tending to slide down); 7.5=gtanθ is the smooth-wedge value; 10 simply equates a to g.
Exam pattern: JEE Main – range of wedge acceleration for no relative sliding (pseudo force with friction)
Question 2Level 3 · AdvancedSingle correct
A simple pendulum of bob mass m hangs from the roof of a car that slides freely down a smooth incline of angle 30∘. In the steady state (bob at rest relative to the car), what are the direction of the string and its tension?
AInclined at 30∘ to the vertical towards the back of the car; tension mgsin30∘
BPerpendicular to the incline; tension mgcos30∘
CPerpendicular to the incline; tension mg
DVertical; tension mg
Show answer and solution
Answer: B
The car's acceleration is gsin30∘ down the incline. In the car frame the pseudo force mgsin30∘ acts up the incline.
Effective gravity =g−a: the component of g along the incline is cancelled, leaving only gcos30∘ perpendicular to the incline.
So the string hangs perpendicular to the incline with T=mgcos30∘.
Trap: "vertical" ignores the car's acceleration; T=mg forgets that the along-slope part of gravity is cancelled.
Exam pattern: JEE Main – pendulum in an accelerating vehicle (effective gravity)
Question 3Level 3 · AdvancedSingle correct
A lift accelerates upward at 2m/s2. Inside it, a block is released from rest at the top of a fixed rough incline of angle 37∘ and length 2.4m; the coefficient of kinetic friction is 0.5. Take g=10m/s2, sin37∘=0.6. How long does the block take to reach the bottom?
A2s
B1s
C1.5s
D2.4s
Show answer and solution
Answer: A
In the lift frame effective gravity is g+a=12m/s2; both the weight component and the normal force (hence friction) scale with it.
Relative acceleration =12(sin37∘−0.5cos37∘)=12(0.6−0.4)=2.4m/s2.
t=2L/a=2×2.4/2.4=2s.
Traps: 2.4 ignores the lift; 1.5 scales the weight component to 12 but leaves friction at μmgcosθ.
Exam pattern: JEE Main – friction on an incline inside an accelerating lift
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